answer：Alright, so I've got this problem about encryption and decryption. It seems like the sender is encrypting some plaintext values into ciphertext using specific rules, and now the receiver has some ciphertext values and needs to figure out the original plaintext. Let me try to understand what's going on here.First, the encryption rules are given as follows: plaintext values (a), (b), and (c) correspond to ciphertext values (a+1), (2b + a), and (3c - 4). They even gave an example where plaintext values 1, 2, 5 correspond to ciphertext values 2, 5, 11. Okay, that makes sense. So, for each plaintext value, there's a specific transformation to get the ciphertext.Now, the receiver has ciphertext values 21, 22, 23, and we need to find the corresponding plaintext values (a), (b), and (c). So, essentially, we need to reverse the encryption process.Let me write down the encryption equations again to get a clear picture:1. ( text{Ciphertext for } a = a + 1 )2. ( text{Ciphertext for } b = 2b + a )3. ( text{Ciphertext for } c = 3c - 4 )Given the ciphertext values are 21, 22, 23. So, we can set up the equations:1. ( a + 1 = 21 )2. ( 2b + a = 22 )3. ( 3c - 4 = 23 )Alright, let's solve these one by one.Starting with the first equation: ( a + 1 = 21 ). To find ( a ), I just need to subtract 1 from both sides.( a = 21 - 1 )( a = 20 )Okay, so ( a ) is 20. That seems straightforward.Now, moving on to the second equation: ( 2b + a = 22 ). We already know ( a = 20 ), so let's substitute that in.( 2b + 20 = 22 )Now, subtract 20 from both sides to isolate the term with ( b ).( 2b = 22 - 20 )( 2b = 2 )Now, divide both sides by 2 to solve for ( b ).( b = frac{2}{2} )( b = 1 )Alright, so ( b ) is 1. That seems pretty simple too.Now, onto the third equation: ( 3c - 4 = 23 ). Let's solve for ( c ).First, add 4 to both sides to get rid of the negative 4.( 3c = 23 + 4 )( 3c = 27 )Now, divide both sides by 3 to solve for ( c ).( c = frac{27}{3} )( c = 9 )So, ( c ) is 9.Let me just recap what I've done:1. For ( a ), I took the first ciphertext value, which was 21, and subtracted 1 to get ( a = 20 ).2. For ( b ), I used the second ciphertext value, 22, and knew that it was equal to ( 2b + a ). Since I already found ( a = 20 ), I substituted that in and solved for ( b ), getting ( b = 1 ).3. For ( c ), I took the third ciphertext value, 23, and knew it was equal to ( 3c - 4 ). I added 4 to both sides and then divided by 3 to find ( c = 9 ).Let me double-check my work to make sure I didn't make any mistakes.Starting with ( a ):( a + 1 = 21 )( a = 20 )Yes, that seems correct.For ( b ):( 2b + a = 22 )( 2b + 20 = 22 )( 2b = 2 )( b = 1 )That also looks correct.For ( c ):( 3c - 4 = 23 )( 3c = 27 )( c = 9 )Yep, that checks out too.So, putting it all together, the plaintext values are ( a = 20 ), ( b = 1 ), and ( c = 9 ).Just to be thorough, let me plug these back into the original encryption rules to see if I get the ciphertext values 21, 22, 23.1. ( a + 1 = 20 + 1 = 21 ) – Correct.2. ( 2b + a = 2(1) + 20 = 2 + 20 = 22 ) – Correct.3. ( 3c - 4 = 3(9) - 4 = 27 - 4 = 23 ) – Correct.Everything matches up, so I feel confident that my solution is correct.I think I've got it. The plaintext values corresponding to the ciphertext 21, 22, 23 are 20, 1, and 9.

answer：Okay, so I've got these three problems about an online store selling Bing Dwen Dwen keychains during the 2022 Beijing Winter Olympics. They bought two types, A and B, with different purchase and selling prices. Let me try to figure out each part step by step.Starting with part (1): They spent 850 yuan to buy a total of 30 keychains of types A and B. I need to find out how many of each type they bought. Alright, so I think I can set up a system of equations here. Let me denote the number of A keychains as x and the number of B keychains as y. From the problem, I know two things:1. The total number of keychains is 30, so x + y = 30.2. The total cost is 850 yuan. Since A costs 30 yuan each and B costs 25 yuan each, the total cost equation would be 30x + 25y = 850.Now, I can solve this system of equations. Maybe I'll use substitution or elimination. Let's try elimination. If I multiply the first equation by 25, I get 25x + 25y = 750. Then, subtract this from the second equation:30x + 25y = 850- (25x + 25y = 750)-------------------5x = 100So, 5x = 100 means x = 20. Then, plugging back into x + y = 30, y = 10. Okay, so they bought 20 A keychains and 10 B keychains. That seems straightforward.Moving on to part (2): After selling all the first round keychains, they plan to buy a total of 80 keychains in the second round, with the total purchase price not exceeding 2200 yuan. They want to maximize the sales profit. First, let's note the profit per keychain. For A, it's 45 - 30 = 15 yuan. For B, it's 37 - 25 = 12 yuan. So, A gives a higher profit per unit.They want to buy 80 keychains in total, so let me denote the number of A keychains as m and B as (80 - m). The total cost should be ≤ 2200 yuan. So, 30m + 25(80 - m) ≤ 2200.Let me compute that:30m + 2000 - 25m ≤ 22005m + 2000 ≤ 22005m ≤ 200m ≤ 40So, they can buy at most 40 A keychains. Since A gives higher profit, to maximize profit, they should buy as many A as possible, which is 40, and the rest 40 would be B.Calculating the profit: 40*15 + 40*12 = 600 + 480 = 1080 yuan. So, maximum profit is 1080 yuan.Now, part (3): They want to adjust the selling price of B keychains. Originally, selling at 37 yuan gives an average of 4 sold per day. For each 1 yuan reduction, they can sell 2 more per day. They want the daily sales profit to be 90 yuan.Let me denote the reduction in price as x yuan. So, the new selling price is (37 - x) yuan. The number sold per day would be 4 + 2x.The profit per keychain is (selling price - purchase price) = (37 - x - 25) = (12 - x) yuan.Total daily profit is (12 - x)*(4 + 2x) = 90.So, setting up the equation:(12 - x)(4 + 2x) = 90Let me expand this:12*4 + 12*2x - x*4 - x*2x = 9048 + 24x - 4x - 2x² = 9048 + 20x - 2x² = 90Let's rearrange:-2x² + 20x + 48 - 90 = 0-2x² + 20x - 42 = 0Multiply both sides by -1 to make it positive:2x² - 20x + 42 = 0Divide all terms by 2:x² - 10x + 21 = 0Now, solving this quadratic equation:x = [10 ± sqrt(100 - 84)] / 2x = [10 ± sqrt(16)] / 2x = [10 ± 4] / 2So, x = (10 + 4)/2 = 7 or x = (10 - 4)/2 = 3So, x can be 7 or 3. That means the selling price can be 37 - 7 = 30 yuan or 37 - 3 = 34 yuan.Let me verify:For x = 7:Price = 30 yuanSold per day = 4 + 2*7 = 18Profit per keychain = 30 - 25 = 5Total profit = 5*18 = 90 yuan. Correct.For x = 3:Price = 34 yuanSold per day = 4 + 2*3 = 10Profit per keychain = 34 - 25 = 9Total profit = 9*10 = 90 yuan. Correct.So, both prices work.I think that's all. Let me just recap:1. First purchase: 20 A and 10 B.2. Second purchase: 40 A and 40 B for max profit of 1080 yuan.3. Selling price for B: Either 30 yuan or 34 yuan to get 90 yuan daily profit.Yeah, that seems to cover all parts.

answer：Okay, so I have this problem here about a telescope increasing the visual range. Let me try to understand what it's asking. It says that a certain telescope increases the visual range at a particular location from some distance to 150 kilometers. The visual range is increased by 50% by using the telescope. I need to find out what the initial visual range was without the telescope.Alright, let's break this down. The telescope increases the visual range by 50%. That means if the original range was, say, X kilometers, then with the telescope, it becomes X plus 50% of X, which is 1.5 times X. And we know that with the telescope, the range is 150 kilometers. So, 1.5 times the original range equals 150 kilometers.Wait, let me make sure I'm interpreting this correctly. If the telescope increases the range by 50%, does that mean the new range is 150% of the original range? Yes, that makes sense because increasing something by 50% means you're adding half of it to itself. So, the new range is 1.5 times the original.So, mathematically, I can write this as:1.5 * V = 150 kmWhere V is the original visual range without the telescope. To find V, I need to solve for it. That means I should divide both sides of the equation by 1.5.V = 150 km / 1.5Let me do that division. 150 divided by 1.5. Hmm, 1.5 goes into 150 how many times? Well, 1.5 times 100 is 150, so V must be 100 kilometers.Wait, let me double-check that. If the original range was 100 kilometers, and the telescope increases it by 50%, then 50% of 100 is 50, so adding that to 100 gives 150 kilometers. Yes, that matches what the problem said.So, the initial visual range without the telescope was 100 kilometers.Just to make sure I didn't make any mistakes, let me think through it again. The telescope adds 50% to the original range, making it 150 kilometers. So, the original range must be less than 150. If I take 150 and divide it by 1.5, I get 100, which seems right because 100 plus 50% of 100 is 150.I think that's solid. There's no need to complicate it further. The key was understanding that a 50% increase means multiplying the original by 1.5, and then reversing that to find the original value.

answer：Okay, so I have this problem here where I'm given three positive real numbers a, b, and c. The question is about the three expressions: a + 1/b, b + 1/c, and c + 1/a. I need to figure out which of the options A, B, C, or D is correct regarding these expressions.Let me first understand the options:A: All three expressions are not greater than 2. So, each of them is ≤ 2.B: All three expressions are not less than 2. So, each of them is ≥ 2.C: At least one of the expressions is not less than 2. So, at least one is ≥ 2.D: At least one of the expressions is not greater than 2. So, at least one is ≤ 2.I need to determine which of these statements must be true for any positive real numbers a, b, and c.Hmm, okay. Maybe I can start by testing some specific values for a, b, and c to see which options hold or don't hold.Let me try option A first. If I can find a counterexample where at least one expression is greater than 2, then A is incorrect.Let's choose a = 3, b = 1, and c can be any positive number, say c = 1.Then, a + 1/b = 3 + 1/1 = 4, which is greater than 2. So, this shows that A is not necessarily true because I found a case where one of the expressions is greater than 2. So, A is incorrect.Now, option B: All three expressions are not less than 2, meaning each is ≥ 2.Again, let's try some specific values. Let me choose a = 1, b = 2, and c = 1.Then, a + 1/b = 1 + 1/2 = 1.5, which is less than 2. So, this shows that B is not necessarily true because I found a case where one of the expressions is less than 2. So, B is incorrect.Moving on to option C: At least one of the expressions is not less than 2, meaning at least one is ≥ 2.I need to see if this is always true. Maybe I can assume the opposite and see if it leads to a contradiction.Assume that all three expressions are less than 2. So,a + 1/b < 2,b + 1/c < 2,c + 1/a < 2.If I add all three inequalities together, I get:(a + 1/b) + (b + 1/c) + (c + 1/a) < 6.Simplify the left side:a + b + c + (1/a + 1/b + 1/c) < 6.Now, I recall that for any positive real number x, x + 1/x ≥ 2 by the AM-GM inequality. So, applying this to each of a, b, c:a + 1/a ≥ 2,b + 1/b ≥ 2,c + 1/c ≥ 2.Adding these three inequalities:(a + 1/a) + (b + 1/b) + (c + 1/c) ≥ 6.But from our earlier assumption, we have:a + b + c + 1/a + 1/b + 1/c < 6.This contradicts the fact that a + b + c + 1/a + 1/b + 1/c is actually ≥ 6. Therefore, our assumption that all three expressions are less than 2 must be false. So, at least one of the expressions must be ≥ 2. Therefore, option C is correct.Let me check option D just to be thorough. Option D says that at least one of the expressions is not greater than 2, meaning at least one is ≤ 2.I need to see if this is always true. Let me try to find a counterexample where all three expressions are greater than 2.Let me choose a = b = c = 2.Then, a + 1/b = 2 + 1/2 = 2.5,b + 1/c = 2 + 1/2 = 2.5,c + 1/a = 2 + 1/2 = 2.5.All three expressions are 2.5, which is greater than 2. So, this shows that D is not necessarily true because I found a case where all expressions are greater than 2. Therefore, D is incorrect.So, after checking all options, only option C is correct.

answer：First, calculate the total distance Christina walks to and from school from Monday to Thursday. She walks 7 km to school and 7 km back home each day, totaling 14 km per day. Over 4 days, this amounts to 56 km.Next, determine the distance Christina covered on Friday. The total weekly distance is 74 km, so subtract the distance covered from Monday to Thursday: 74 km - 56 km = 18 km.On Friday, Christina walks the usual 14 km to and from school. The additional 4 km she walked indicates the distance to her mother's friend's place.Therefore, the distance from the school to her mother's friend's place is 4 km.

answer：First, I need to simplify the expression ((2a^{2}b)^{3}).I recall that when raising a product to a power, each factor inside the parentheses is raised to that power individually.So, I'll apply the exponent of 3 to each part of the expression:1. (2^{3} = 8)2. ((a^{2})^{3} = a^{6})3. (b^{3} = b^{3})Now, I'll combine these results:(8 cdot a^{6} cdot b^{3} = 8a^{6}b^{3})Therefore, the simplified form of the expression is (8a^{6}b^{3}).